Solve: 20 + 6 ÷ [(51 ÷ C) - (5 × C)] - Step-by-Step Guide

Hey there, math enthusiasts! Today, we're diving into an intriguing equation that might seem a bit daunting at first glance. But don't worry, we'll break it down step by step, making sure everyone can follow along. Our mission? To find the value of $20 + 6 \div [(51 \div (C) - (5 \times (C)]$. Get ready to put on your thinking caps and let's get started!

Decoding the Equation: A Step-by-Step Approach

1. Understanding the Order of Operations

Before we jump into solving, let's quickly revisit the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). This golden rule tells us the sequence in which we should tackle the different parts of the equation. It's like the secret code to unlocking the solution! Ignoring this order will lead to wrong answers, and we definitely don't want that. Imagine trying to build a house without a blueprint - that's what solving math without PEMDAS is like. It is important to solve parentheses first, followed by exponents, then multiplication and division (from left to right), and finally addition and subtraction (also from left to right). In our equation, we've got parentheses, division, multiplication, and addition. So, let's keep PEMDAS in our minds as we move forward.

2. Focusing on the Parentheses: The Inner Circle

Our equation features a rather prominent set of parentheses: $[(51 \div (C) - (5 \times (C)]$. This is where the real action begins. Inside these parentheses, we find a mix of division and multiplication, both involving the mysterious variable 'C'. To crack this part, we need to address the operations within the inner parentheses first. Remember, parentheses are like the VIP section of the equation – they get our exclusive attention first! The expression inside the parentheses is what we need to simplify before we can move on to the rest of the equation. This might seem like a small part of the problem, but it's a crucial step. We're essentially setting the stage for the rest of our calculations. So, let's dive into the inner workings of these parentheses and see what we can uncover. Remember, the 'C' is our unknown variable, and how we handle it within this section will determine the course of our solution. This is where the puzzle starts to take shape, and we're one step closer to finding the answer.

3. Tackling Division and Multiplication Inside the Parentheses

Inside our parentheses, we have two key operations vying for our attention: division and multiplication. Specifically, we're looking at $51 \div (C)$ and $5 \times (C)$. According to PEMDAS, multiplication and division hold equal importance, so we tackle them from left to right. Think of it as a race – the one that appears first gets our attention first! So, we first handle $51 \div (C)$, which can also be written as $\frac{51}{C}$. This is a crucial step because it starts to simplify the expression within the parentheses. Next, we move on to $5 \times (C)$, which is simply $5C$. This is straightforward multiplication, and it's the second piece of the puzzle within our parentheses. Now, we have two simplified terms, $\frac{51}{C}$ and $5C$, which we will use in the next step. This is like assembling the individual parts of a machine before putting them together. Each part has its role, and together, they will help us solve the equation. Remember, keeping track of each step is essential in math, just like following a recipe when you're cooking. So, let's move on to the next step and see how these pieces fit together.

4. Combining Terms Within the Parentheses

Now that we've handled the division and multiplication, we can rewrite the expression inside the parentheses as $\frac{51}{C} - 5C$. This looks a bit cleaner, doesn't it? We've transformed the original expression into a more manageable form. To combine these terms, we need a common denominator, which in this case is 'C'. Think of it like adding fractions – you can't add them directly unless they have the same denominator. So, we rewrite $5C$ as $\frac{5C^2}{C}$. Now our expression looks like $\frac{51}{C} - \frac{5C^2}{C}$. This is a critical step because it allows us to combine the two terms into a single fraction. Now we can subtract them, giving us $\frac{51 - 5C^2}{C}$. This single fraction represents the simplified form of the expression inside the parentheses. We've taken a complex expression and condensed it into a more compact form. This is a significant milestone in our journey to solve the equation. Now, with the parentheses simplified, we can move on to the next operation in our equation. Remember, math is like a puzzle, and each step brings us closer to the final picture.

5. Tackling the Division Outside the Parentheses

With the parentheses simplified to $\frac{51 - 5C^2}{C}$, we can now address the division outside the parentheses. Our equation now looks like $20 + 6 \div \frac{51 - 5C^2}{C}$. Remember, dividing by a fraction is the same as multiplying by its reciprocal. It's like flipping the fraction upside down and then multiplying. So, $6 \div \frac{51 - 5C^2}{C}$ becomes $6 \times \frac{C}{51 - 5C^2}$, which simplifies to $\frac{6C}{51 - 5C^2}$. This step is a game-changer because it transforms a division problem into a multiplication problem, which is often easier to handle. We've essentially performed a mathematical maneuver that makes our equation more approachable. Now, we have a single fraction that represents the result of the division. This is a crucial piece of the puzzle, and it sets us up for the final step: addition. Remember, in math, as in life, sometimes you have to change your perspective to find the best path forward. And that's exactly what we've done here by converting division into multiplication. Let's move on to the final step and bring it all home!

6. The Final Step: Adding 20 to the Result

We've reached the final stretch! Our equation has been simplified to $20 + \frac{6C}{51 - 5C^2}$. Now, we just need to add 20 to the fraction. To do this, we need to express 20 as a fraction with the same denominator as $\frac{6C}{51 - 5C^2}$. This means we need to rewrite 20 as $\frac{20(51 - 5C^2)}{51 - 5C^2}$. Think of it as finding a common language for the two terms so they can communicate and combine. When we distribute the 20, we get $\frac{1020 - 100C^2}{51 - 5C^2}$. Now we can add the two fractions: $\frac{1020 - 100C^2}{51 - 5C^2} + \frac{6C}{51 - 5C^2}$. This gives us a single fraction: $\frac{1020 - 100C^2 + 6C}{51 - 5C^2}$. We've successfully combined all the terms into one fraction! This is the culmination of all our efforts, the grand finale of our mathematical journey. We've taken a complex equation and broken it down step by step, and now we have a single, simplified expression. This is a moment of triumph, like reaching the summit of a mountain after a long climb. But remember, our journey isn't over until we find the value of 'C'. So, let's hold onto this result and see what we need to do next to solve for 'C'.

Putting It All Together: Finding the Value

To find the value, we need the value of 'C' . Without knowing the value of 'C', we can't get a numerical answer. This is where the problem might be incomplete, or we're missing some crucial information. But hey, we've done the hard work of simplifying the equation. That's a victory in itself!

Wrapping Up: What We've Learned

So, what have we learned today? We've tackled a complex equation, navigated the order of operations, and simplified a tricky expression. We've seen the power of PEMDAS and the importance of breaking down problems into smaller, manageable steps. We've also learned that sometimes, we need more information to reach the final answer. But even without that final answer, we've gained valuable skills and knowledge. Math is like a journey, and every step we take makes us stronger and more confident. Keep practicing, keep exploring, and keep those math muscles flexing!